How to find $\int_0^1 x^4(1-x)^5dx$ quickly? This question came in the Rajshahi University admission exam 2018-19
Q) $\int_0^1 x^4(1-x)^5dx$=?
(a) $\frac{1}{1260}$
(b) $\frac{1}{280}$
(c)$\frac{1}{315}$
(d) None
This is a big integral (click on show steps):
$$\left[-\dfrac{\left(x-1\right)^6\left(126x^4+56x^3+21x^2+6x+1\right)}{1260}\right]_0^1=\frac{1}{1260}$$
It takes a lot of time to compute. How can I compute this quickly (30 seconds) using a shortcut?
 A: Maybe it is not the fastest way to solve it, but it is faster than computing the original integral:
\begin{align*}
I = \int_{0}^{1}x^{4}(1 - x)^{5}\mathrm{d}x = \int_{0}^{1}(1 - x)^{4}x^{5}\mathrm{d}x & \Rightarrow 2I = \int_{0}^{1}x^{4}(1-x)^{4}(x + (1 - x))\mathrm{d}x\\\\
& \Rightarrow 2I = \int_{0}^{1}x^{4}(1-x)^{4}\mathrm{d}x\\\\
& \Rightarrow 2I = \int_{0}^{1}x^{4}(1 - 4x + 6x^{2} - 4x^{3} + x^{4})\mathrm{d}x\\\\
& \Rightarrow 2I = \int_{0}^{1}(x^{4} - 4x^{5} + 6x^{6} - 4x^{7} + x^{8})\mathrm{d}x\\\\
& \Rightarrow 2I = \frac{1}{5} - \frac{2}{3} + \frac{6}{7} - \frac{1}{2} + \frac{1}{9}\\\\
& \Rightarrow I = \frac{1}{1260}
\end{align*}
A: If you don't have the beta function formula memorized, you can use integration by parts repeatedly
\begin{align*}
\int_{0}^{1}x^4(1-x)^5\,dx &= \left[x^4 \cdot -\dfrac{1}{6}(1-x)^6\right]_{0}^{1} - \int_{0}^{1}4x^3 \cdot -\dfrac{1}{6}(1-x)^6\,dx
\\
&= \dfrac{4}{6}\int_{0}^{1}x^3(1-x)^6\,dx
\\
&= \dfrac{4}{6}\left[x^3 \cdot -\dfrac{1}{7}(1-x)^7\right]_{0}^{1} - \dfrac{4}{6}\int_{0}^{1}3x^2 \cdot -\dfrac{1}{7}(1-x)^7\,dx
\\
&= \dfrac{4 \cdot 3}{6 \cdot 7}\int_{0}^{1}x^2(1-x)^7\,dx
\\
&= \dfrac{4 \cdot 3}{6 \cdot 7}\left[x^2 \cdot -\dfrac{1}{8}(1-x)^8\right]_{0}^{1} - \dfrac{4 \cdot 3}{6 \cdot 7}\int_{0}^{1}2x \cdot -\dfrac{1}{8}(1-x)^8\,dx
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8}\int_{0}^{1}x(1-x)^8\,dx
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8}\left[x \cdot -\dfrac{1}{9}(1-x)^9\right]_{0}^{1} - \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8}\int_{0}^{1}1 \cdot -\dfrac{1}{9}(1-x)^9\,dx
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8 \cdot 9}\int_{0}^{1}(1-x)^9\,dx
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8 \cdot 9}\left[-\dfrac{1}{10}(1-x)^{10}\right]_{0}^{1}
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8 \cdot 9 \cdot 10}
\\
&= \dfrac{1}{1260}
\end{align*}
This looks fairly long since I wrote out all the steps in detail. However, if you notice the pattern, you can probably arrive at the answer faster. Also, if are familiar with the Tabular Integration Method, then you only need to do a fairly short amount of scratchwork.
A: $$I=\int_{0}^{1}x^4(1-x)^5 dx$$
Let $x=\sin^2 t$ then
$$I=\int_{0}^{\pi/2} 2 \sin^9 t \cos^{11} t dt$$
Use Beta integral
$$\int_{0}^{\pi/2} \sin^{x}t \cos^{y}t dt=\frac{1}{2} \frac{\Gamma((x+1)/2)\Gamma((y+1)/2)}{\Gamma((x+y+2)/2)}$$
to get
$$I=\frac{\Gamma(5)\Gamma(6)}{\Gamma(11)}=\frac{4! 5!}{10!}=\frac{1}{1260}$$
A: Personally I'm somewhat inclined to use the fifth row of Pascal's triangle: $$  
                 1\\
                1\quad 1 \\
               1\quad  2\quad 1\\
              1\quad  3\quad 3\quad 1\\
            1 \quad 4\quad  6\quad  4 \quad 1\\
            1\quad  5\quad  10\quad 10\quad 5 \quad 1$$
Thus $$(1-x)^5=1-5x^1+10x^2-10x^3+5x^4-x^5$$.
So, $$x^4-5x^5+10x^6-10x^7+5x^8-x^9$$ is the integrand.
We get $$ 1/5-5/6+10/7-10/8+5/9-1/10=-19/30+10/56+41/90=-16/90+10/56=-8/45+5/28=1/1260$$.
A: Use Beta function:
$$
B(a,b)=\int_0^1x^{a-1}(1-x)^{b-1}\mathrm dx={\Gamma(a)\Gamma(b)\over\Gamma(a+b)}.
$$
Since $\Gamma(a)=(a-1)!$ when $a$ is a positive integer, we have
$$
\int_0^1x^4(1-x)^5\mathrm dx=B(5,6)={\Gamma(5)\Gamma(6)\over\Gamma(11)}={4!\dot5!\over10!}={4\cdot3\cdot2\over10\cdot9\cdot8\cdot7\cdot6}={1\over1260}.
$$
A: I wouldn't say this method helps you in solving it in 30 seconds, but I think it can help you in simplifying the calculations so that the integral can be computed faster
First consider by usage of the Binomial Theorem
$$(1-x)^5=\sum_{k=0}^{5}\binom{5}{k}(1)^k(-x)^{5-k}=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}(x)^{5-k}$$
from which you can get by multiplying by $x^4$
$$x^4(1-x)^5=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}(x)^{9-k}$$
applying the integral to the sum
$$\int_{0}^{1}x^4(1-x)^5\cdot dx=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}\bigg(\frac{x^{10-k}}{10-k} \bigg)_{0}^{1}$$
which will give
$$\int_{0}^{1}x^4(1-x)^5\cdot dx=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}\bigg(\frac{1}{10-k} \bigg)$$
Expanding the sum gives
$$-\binom{5}{0}\bigg(\frac{1}{10}\bigg)+\binom{5}{1}\bigg(\frac{1}{9}\bigg)-\binom{5}{2}\bigg(\frac{1}{8} \bigg)+\binom{5}{3}\bigg(\frac{1}{7}\bigg)-\binom{5}{4}\bigg(\frac{1}{6} \bigg)+\binom{5}{5}\bigg(\frac{1}{5} \bigg)$$
simplifying will give
$$\frac{-1}{10} +\frac{5}{9} -\frac{10}{8} +\frac{10}{7} -\frac{5}{6} +\frac{1}{5}=\frac{1}{1260}$$
A: We have
$$\begin{align}
\int_{0}^{1}x^4(1-x)^5\,dx
&=\int_{-1/2}^{1/2}(1/2+x)^4(1/2-x)^5\,dx\\
&=\int_{-1/2}^{1/2}(1/4-x^2)^4(1/2-x)\,dx\\
&=\int_{-1/2}^{1/2}(1/4-x^2)^4(1/2)\,dx\\
&\qquad+\int_{-1/2}^{1/2}(1/4-x^2)^4(-x)\,dx\\
&=\int_{-1/2}^{1/2}(1/4-x^2)^4(1/2)\,dx\\
&=\int_{0}^{1/2}(1/4-x^2)^4\,dx,
\end{align}$$
which is easily computed to $1/1260$.
